Đề bài: Cho hàm số \(f\left( x \right) = \frac{{{9^x}}}{{{9^x} + 3}}\). Tính tổng \(S = f\left( {\frac{1}{{2017}}} \right) + f\left( {\frac{2}{{2017}}} \right) + … + f\left( {\frac{{2016}}{{2017}}} \right) + f\left( 1 \right)\)

Lấy a, b sao cho \(a + b = 1,\) ta có:

\(\begin{array}{l}f(a) + f(b) = \frac{{{9^a}}}{{{9^a} + 3}} + \frac{{{9^b}}}{{{9^b} + 3}} = \frac{{{9^a}({9^b} + 3) + {9^b}({9^a} + 3)}}{{({9^a} + 3)({9^b} + 3)}}\\ = \frac{{{9^a}{{.9}^b} + {{39}^a} + {9^b}{{.9}^a} + {{39}^b}}}{{{9^a}{{.9}^b} + {{3.9}^a} + {{3.9}^b} + 9}} = \frac{{{9^1} + {9^1} + 3({9^a} + {9^b})}}{{{9^1} + 3({9^a} + {9^b}) + 9}} = 1.\end{array}\)

Ta có:

 \(\begin{array}{l}S = f\left( {\frac{1}{{2017}}} \right) + f\left( {\frac{2}{{2017}}} \right) + … + f\left( {\frac{{2016}}{{2017}}} \right) + f\left( 1 \right)\\ = f\left( {\frac{1}{{2017}}} \right) + f\left( {\frac{{2016}}{{2017}}} \right) + f\left( {\frac{2}{{2017}}} \right) + f\left( {\frac{{2015}}{{2017}}} \right) + …. + f(1)\\ = 1008 + \frac{3}{4} = \frac{{4035}}{4}.\end{array}\)